Re: quantum mechanics and curvature of space-time

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 09/29/04 

Date: Wed, 29 Sep 2004 07:25:56 +0000 (UTC)

alistair@goforit64.fsnet.co.uk (alistair) wrote in message news:<861c1b21.0409241312.3e02324@posting.google.com>...
> The amplitude of a wavefunction squared is proportional to the
> probability of finding a particle at a position in space.
> This squaring of the amplitude is often compared to the intensity
> (energy density) of an electromagnetic wave.
> The stress energy momentum tensor of general relativity contains
> the term Too - the energy density.Can Too be related to the
> square of the amplitude of the quantum mechanical wavefunction?
> Is Too (and therefore the curvature of spacetime due to energy
> density) proportional to the probability of finding a particle at a
> position in space?

Offhand I would say yes but with an unorthodox caveat,
but I'm curious too.

The usual Einstein Law is G_uv = T_uv (approximately).

When solving that equation near the Sun it is conventional
to set G_uv =0 meaning the energy density is zero and the
point where the solution occurs is regarded as a pure vacuum.

But I ask and question "what is a pure vacuum"?

Do we measure a volume of 1 cubic meter and find no
particles, or should we measure 1,000,000 cubic miles
and find the Sun within that volume and arrive at a
different answer for density?

So the G_uv may very well be a geometric representation
of the probability field, provided we define what a
vacuum is and over what volume we should use to calculate
energy density.
Regards
Ken S. Tucker

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